I thought that was a good one. No, but my aunt has long since given up on discussions with the rest of the family, and I'm at least wise enough to keep my mouth shut and read some Joyce or another.
would the following statement of the problem be correct? (I think i was misreading it initially)
suppose i make a trip from one bank of a river to another, and then back. No matter what route I take, it'll be faster with a river current than without.
@MikeMiller Is it true that if you take a tubular neighborhood of $\Bbb{CP}^1 \subset \Bbb{CP}^2$, then the boundary of that is an $S^3$? 'Cause it's a circle bundle on $S^2$... it seems to be the case visually, I think, but it's not clear. I don't think I know enough to prove it.
Right, almost. I intended a trip down-river and then back up-river, or else the algebra I exercise doesn't apply. For the closed curve case, it could be any closed curve whatsoever.
You and I did stuff like this, @Balarka, ages ago. Take the hyperplane to be $z_0=0$ in $\Bbb P^2$. What does the boundary of a neighborhood look like?
@TedShifrin If $n = 2$, you can look at two transverse copies of $\Bbb P^1$ in $\Bbb P^2$, and homotope their union by perturbing them at the intersection point to change $z_1 z_2 = 0$ to $z_1 z_2 = 1$. This is an embedded sphere representing $2\alpha$. Topologically, this is the same as picking two $D^2$'s with boundary being fibers in the Hopf bundle (hence, linked as a Hopf link), and perturbing a bit to make it a Seifert surface. Then capping off the link components gives an embedded sphere.
@TedShifrin I haven't thought a lot (hence, not comfortable) about the algebro-geometric interpretation of projective space, but let me try to answer that. Do you want me to get a neighborhood of the corresponding hyperplane in $\Bbb C^3$ by using the metric?
These days I try to make it more of a policy (not 100%, but close) to only have friends whom I actually know. Some people came in earlier (e.g., Ramya). Obviously, Kaj and I know each other quite well.
A subset $A \subset \Bbb C$ is called simply connected if given any map $f : [0, 1] \to A$ such that $f(0) = f(1) = x$ there is a continuous sequence of maps $f_t : [0, 1] \to A$ for $t \in [0, 1]$ with $f_1 = f$ and $f_0$ the constant map sending everything to $x$, such that $f_t(0) = f_t(1) = x$ for each $t \in [0, 1]$.
And the sequence $\{f_t\}$ is continuous in the sense that the map $H: [0, 1] \times [0, 1] \to A$ given by $H(s, t) = f_t(s)$ is continuous.
I am not sure if I parse that question. A homotopy is a sequence of loops tending to the constant loop. So, like, you're continuously "tightening" the loop at $x$ (fixing $x$!), and $f_t$ is the trace of that continuous deformation at time $t$ (i.e., that's the image you get when you stop at time $t$).
The above image might help. However, be warned, we're not doing this with subsets of $\Bbb C$ anymore.
This is cool stuff. So if it wasn't simply connected, we'd have our continuously deforming loop falling in an area that's a "hole" of the domain or where the domain isn't defined, @BalarkaSen?
@TedShifrin since the vector field of the water flowing in the river is always parallel to the curve of the river itself, the dot product between the river flow vector and the tangent vector will always be positive assuming some flow exists.
Therefore the line integral around a closed curve with a boat in the river will always be positive, i.e. work is done. Therefore, the initial velocity should be greater than the starting velocity since F = ma and therefore, it has a positive acceleration meaning v is increasing around the curve.
Hi, I am trying to fit an Edgeworth expansion to some non identically distributed random variables (different mean, same variance). Typically Edgeworth expansion is derived assuming i.i.id. variables. Is there a general definition?
Is there a program where I can see the shape of the surface $x^2+y^2+z^4=1$ ? When I write this equation in Wolfram, I don't get the picture of the surface...
We consider a sphere. Does the unit normal of the sphere and the radius vector have always the same direction? Or can it also be that they have opposite directions?
@BalarkaSen: BTW, I asked you about topological group $\implies$ $\pi_1(G)$ abelian the other day. you probably already know this but if $G$ has a universal cover, it becomes very easy to prove the statement.
@Huy It's not completely clear to me how you would use $\tilde{G}$ to prove that the thing's abelian. As I mentioned previously, the proof I am familiar with uses Eckmann-Hilton argument.
oh, sure, I wasn't sure if you knew the argument yet so I mentioned it just in case you didn't!
let $G$ be a connected topological group. you can prove very quickly that if $N \unlhd G$ is a discrete, normal subgroup, then it is contained in the center of $G$.
@Huy Ok, I see. But to do that you need to prove a buckload of things, i.e., prove $\tilde{G}$ is a group, the universal covering map $p$ is actually a morphism of groups, etc. :)
So I am not sure if I prefer this argument over Eckmann-Hilton (which is like 1 line).
Thanks though, I wasn't familiar with this argument.
@Huy Really? If $G$ is a topological group, you have a binary operation $G \times G \to G$ and the inversion map $i : G \to G$. When applied $\pi_1$, you get corresponding binary op $\otimes : \pi_1(G) \times \pi_1(G) \cong \pi_1(G \times G) \to \pi_1(G)$ and inversion $i_* : \pi_1(G) \to \pi_1(G)$. So now you have two group structures on $\pi_1(G)$ - one, $*$ (loop mult) and another $\otimes$.
You can show that these two commutes with each other and conclude that they are the same from there (Eckman-Hilton, this is explicitly done in an exercise of Munkres' Topology).
See, the deal is, you have one group structure on $\pi_1(G)$ (the usual). But you also have the following group structure $\otimes : \pi_1(G) \times \pi_1(G) \to \pi_1(G)$ defined as $[f] \otimes [g] = [f \cdot g]$ where $f \cdot g : [0, 1] \to G$ is defined as $s \mapsto f(s) \cdot g(s)$ (multiply each pt in the two curves which you can do as $G$ is a group).
is it obvious that this is well-defined? seems to me like there could be something going wrong if I multiply with some different representative (of the identity, for example)
(I feel like it can't go wrong due to continuity of multiplication but I'm not sure how to put it into a rigorous argument)
@BalarkaSen: "In its original meaning, jealousy is distinct from envy, though the two terms have popularly become synonymous in the English language, with jealousy now also taking on the definition originally used for envy alone."
@MikeMiller Ok, but I was just saying that E-H proves a much more general statement about H-spaces quite elementarily. I don't see how that statement about $\pi_n$ is relevant.
Should you see this @TedShifrin, could you let me know which program you used to create the awesome diagrams from your Differential Geometry notes? Thanks in advance! :-)
Let $\gamma$ be a unit-speed curve of the sphere of center $\alpha$ and radius $r$, then the unit normal of the sphere is $\textbf{N}=\pm \frac{\gamma-\alpha}{\|\gamma-\alpha\|}$, right? @robjohn
@MaryStar it doesn't matter the speed of the curve, $\frac{\gamma-\alpha}{\|\gamma-\alpha\|}$ is the outward unit normal, and $-\frac{\gamma-\alpha}{\|\gamma-\alpha\|}$ is the inward unit normal.
@robjohn Yes, I mean the general formula for the normals of a sphere.
So I had to say:
The unit normals of the sphere (inward and outward normal) have the same direction with the radius vector, so we have $\textbf{N}=\pm \frac{\gamma-\alpha}{\|\gamma-\alpha\|}$.
I wonder if anyone could kindly help me with a confusion. Consider an $n$ by $n$ binary matrix and a random vector $v$ whose elements are randomly chosen from $\{0,1\}$. We know that the entropy of MV, that is $H(Mv)$ is $n$ as long as $M$ is non-singular.
Oh that's awesome, @TedShifrin. Thanks! (Reason I'm asking is because I wanna write an exposition on harmonic functions/spherical functions and their orthogonal representations so will probably need to include diagrams!)
it is ok for the most part like the only things you can do here is watch shows, drink or gamble, or partying. I don't like drinking,gambling or partying and shows are expensive
Hm, I'm not familiar with the history of these things, but I'm wondering: did one introduce smash to imitate the tensor product or was it "Ok, guys, we really need something so that $S^n * S^m = S^{n+m}$"
@MikeM: You need to work through all the stuff on complexification of tangent bundles of complex manifolds. When you're thinking of the ${\scr O}_X$ module, that is the holomorphic tangent bundle, which is what $T^{(1,0)}X$ is when you decompose $T^*X\otimes\Bbb C$.
I do not know the motivation, but you can use smash product to give $[X, K(G, n)]$ a ring structure. That is, a map $[X, K(G, n)] \times [X, K(G, m)] \to [X, K(G, n+m)]$. (of course, $[X, K(G, n)]$ also has a group structure as $K(G, n)$ is an H-space, so that ring structure is actually the cup product in cohomology).
@Ted: But I think of vector bundles in a non-algebrwic way: cocycles. I don't think I can do that for things other than line bundles on varieties, can I?
